smuklica8i

2022-07-22

The theorem is stated as follows:

Assume that $f:\left[a,b\right]\to \mathbb{R}$ is continuous and that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs. Then there is a point $c\in \left(a,b\right)$ such that $f\left(c\right)=0$.

The beginning of the proof is as follows:

Consider the case where $f\left(a\right)<0. Let $A=\left\{x\in \left[a,b\right]:f\left(x\right)<0\right\}$ and set $c=supA$. Observe first that since $f$ is continuous and $f\left(b\right)>0$, then $c.

I don't see why $c follows from the fact that f is continuous and $f\left(b\right)>0$, but I would really like to understand this, especially since the author seems to think it needs no further explanation. I did try looking at other proofs for the IMT, on this site and other sites, but none of them phrased the proof in this way and they didn't really further my understanding.

abortargy

Expert

If $f$ is continous then there exists a ball with the center in $b$ such that $f\left(x\right)>0$0 in this ball. We could take for example the ball of the radius $\epsilon$ such that

Such $\epsilon$ exists according to continuity of $f$. So, then $c\le b-\epsilon

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Recalculate according to your conditions!